Chapter 9, Problem 49RE

a.

To determine

Find the standard form of the equation of the hyperbola.

Answer to Problem 49RE

The standard form of equation of the hyperbola is $\frac{\text{(x-1)²}}{16}-\frac{\text{(y+1)²}}{9}=1$ .

Explanation of Solution

Given:

It is given in the question that the equation is $\text{9x² - 16y² - 18x + 32y - 151 = 0}$ .

Concept Used:

In this, use the concept of dividing the Right Hand term to Left hand term make the equation like this, $\frac{x^2}{{(a)}^2}-\frac{y^2}{{(b)}^2}=1$ .

Calculation:

Here, the equation is $\text{9x² - 16y² - 18x + 32y - 151 = 0}$ .

Break the term in factor-ise form:

  $\begin{array}{l}\text{9x² - 16y² - 18x - 32y - 151 = 0}\\ \text{=> 9x² - 18x - 16y² - 32y - 151 = 0}\\ \text{=>9(x²-2x) -16(y²+2y) =151}\\ \text{=>9(x²-2x + 1 -1) - 16(y²+2y+1 -1) =151}\\ \text{=>9(x-1)²-9 -16(y+1)² + 16 = 151}\\ \Rightarrow \text{9(x-1)² -16(y+1)² = 144}\end{array}$

Divide the RHS term to the LHS side and make RHS one.

  $\begin{array}{l}\text{9(x-1)² -16(y+1)² = 144}\\ \frac{\text{(x-1)²}}{16}-\frac{\text{(y+1)²}}{9}=1\end{array}$

Conclusion:

The standard form of equation is $\frac{\text{(x-1)²}}{16}-\frac{\text{(y+1)²}}{9}=1$ .

b.

To determine

Calculate the center, vertices, foci and eccentricity of the hyperbola.

Answer to Problem 49RE

The centre, vertices, foci and eccentricity are $(1,-1);(5,-1),(-3,-1);(6,-1),(-4,-1);\frac{5}{4}$

Explanation of Solution

Given: It is given in the question that the standard form is $\frac{\text{(x-1)²}}{16}-\frac{\text{(y+1)²}}{9}=1$ .

Concept Used:

In this, use the concept of finding the centre, vertices by $(h,k\pm a)$ ,foci by $(h,k\pm c)$ and eccentricity by $e=\frac{c}{a}$ by understanding the standard form.

Calculation:

Here, the standard form is $\frac{\text{(x-1)²}}{16}-\frac{\text{(y+1)²}}{9}=1\Rightarrow \frac{\text{(x-1)²}}{{(4)}^2}-\frac{\text{(y+1)²}}{{(3)}^2}=1$ .

Now, the centre is $(1,-1)$ .

Again, horizontal transverse axis , $a=4,b=3$ .

The vertices of the hyperbola is $(h\pm a,k)=(5,-1),(-3,-1)$ .

Now, use a and b to find c.

  $c=\sqrt{a^2+b^2}=\sqrt{{(4)}^2+{(3)}^2}=\sqrt{16+9}=5$

Then, foci of the hyperbola is $(h\pm c,k)=(6,-1),(-4,-1)$ .

At last, the eccentricity of the hyperbola,

  $\begin{array}{l}e=\frac{c}{a}\\ e=\frac{5}{4}\end{array}$

Conclusion:

The centre, vertices, foci and eccentricity are $(1,-1);(5,-1),(-3,-1);(6,-1),(-4,-1);\frac{5}{4}$

c.

To determine

Draw the graph of the hyperbola.

Explanation of Solution

Given:

It is given in the question that the standard form is $\frac{\text{(x-1)²}}{16}-\frac{\text{(y+1)²}}{9}=1$ .

Graph:

  

Interpretation:

Here, the standard form of equation is $\frac{\text{(x-1)²}}{16}-\frac{\text{(y+1)²}}{9}=1$ , Put that equation in the graphing calculator , it will give the required graph in a proper manner.